Question: Find the point on the line
\[y = -3x + 5\]that is closest to the point $(-4,-2).$
Answer: Note that $(0,5)$ and $(1,2)$ are two points on the line, so the line has a direction vector of
\[\begin{pmatrix} 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}.\][asy]
unitsize(0.4 cm);

pair A, B, C, D, P, V;

A = ((5 + 10)/3, -10);
B = ((5 - 10)/3, 10);
V = (-4,-2);
P = (V + reflect(A,B)*(V))/2;
C = (0,5);
D = (1,2);

draw((-10,0)--(10,0));
draw((0,-10)--(0,10));
draw(A--B,red);
draw(V--P,dashed);
draw(C--V,Arrow(6));
draw(C--D,Arrow(6));

dot("$(-4,-2)$", V, SW);
dot("$(0,5)$", C, E);
dot("$(1,2)$", D, E);
[/asy]

The vector going from $(0,5)$ to $(-4,-2)$ is $\begin{pmatrix} -4 \\ -2 \end{pmatrix} - \begin{pmatrix} 0 \\ 5 \end{pmatrix} = \begin{pmatrix} -4 \\ -7 \end{pmatrix}.$  Projecting this vector onto the direction vector, we get
\[\operatorname{proj}_{\begin{pmatrix} 1 \\ -3 \end{pmatrix}} \begin{pmatrix} -4 \\ -7 \end{pmatrix} = \frac{\begin{pmatrix} -4 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \end{pmatrix}}{\left\| \begin{pmatrix} 1 \\ -3 \end{pmatrix} \right\|^2} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \frac{17}{10} \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} \frac{17}{10} \\ -\frac{51}{10} \end{pmatrix}.\][asy]
usepackage("amsmath");

unitsize(0.4 cm);

pair A, B, C, D, P, V;

A = ((5 + 10)/3, -10);
B = ((5 - 10)/3, 10);
V = (-4,-2);
P = (V + reflect(A,B)*(V))/2;
C = (0,5);
D = (1,2);

draw((-10,0)--(10,0));
draw((0,-10)--(0,10));
draw(A--B,red);
draw(V--P,dashed);
draw(C--V,Arrow(6));
draw(C--P,Arrow(6));

dot("$(-4,-2)$", V, SW);
dot("$(0,5)$", C, E);
dot("$\begin{pmatrix} \frac{17}{10} \\ -\frac{51}{10} \end{pmatrix}$", P, NE);
[/asy]

Then
\[\begin{pmatrix} 0 \\ 5 \end{pmatrix} + \begin{pmatrix} \frac{17}{10} \\ -\frac{51}{10} \end{pmatrix} = \begin{pmatrix} \frac{17}{10} \\ -\frac{1}{10} \end{pmatrix},\]so the point on the line closest to $(-4,-2)$ is $\boxed{\left( \frac{17}{10}, -\frac{1}{10} \right)}.$